In Exercises 1 - 24, use Gaussian Eliminaion to find the complete...

Question

In Exercises 1 - 24, use Gaussian Eliminaion to find the complete solution to each system of equations, or show that none exists.5x + 12y + z = 102x + 5y + 2z = - 1x + 2y - 3z = 5

Accepted Answer

Welcome back. I am so glad you're here. We're given this system of equations and we're asked to solve for it using Gaussian elimination. Will we recall from previous lessons that when we have a system of equations, we can create a matrix using the coefficients and the constant terms from our system of equations. So for the first equation we can copy down the coefficients and the constant term and that will be the first row of our matrix. So the first rover matrix is 15, 36 3 and 30. The second row of our matrix is going to come from the second equation. So that will be 4 10, 4 negative two. The third rover matrix will be from the third equation negative 9 15. And what we want is to get this, get this matrix with a diagonal row of ones in it. So starting in the upper left hand corner we're going to have a one and then it will have zeros in the column below it. And then the second column will have something and then a one and a zero below it. The third column will have something and something else and then the one cr diagonal row of ones. And then the fourth column will be something something and then our final position will be our Z. Value. And once we've solved for Z, then we can go back in and substitute and sulfur Y. And then once we sulfur Y we can solve for X. So in order to get all of that done, we are going to start off by getting this upper left hand position. This 15 turned into a one and we're going to replace rho one. How do we get that? 15 to be a one? Well, we can take row one and divide everything by 15 or multiply it by 1/15. So we're going to take row one times 1/15. Because we're replacing row one, we can copy down Rose two and three. So that's 4 10, 4 negative negative 9 15. And our new row one will be 15 times 1/15 is the one that we want. 36 times 1/15 is going to be 5th. So three times 1 15th is 1 5th and 30 times 1, 15 is too Okay, now we're going to continue working on the first column and we want a zero below that one. Excuse me? How do we get that? Four to become a zero? Well, we're going to work with the row that has the one in the desired position. So that's rho one. In order to replace row two, we're going to take row one, multiply it by negative four And add that to row two. Because we're replacing road to we can copy over Rose one and three. So that'll be one 12 5th 1/ 236, negative 9 15. Now replacing row two negative four times one is negative four plus four is zero, negative four times 12, 5th is negative 48 5th plus is 2/5 negative four times one. Fifth is negative 4/5 plus four is 5th and negative four times two is negative eight plus negative two is negative 10. Okay, now we want to continue with that column and get a zero in the final position. So we'll be replacing row three. How do we get that three to become a zero? We're going to work with the row that has the one in the desired position. So row one, we'll take row one multiplied by negative three and then add that to row three. Because we are replacing row three, we can copy down rows one and two. So 1 12 5th. So 1/5 20, 2/5 16 5th, so negative 10. And now doing the work to replace row three negative three times one is negative three plus three is the zero that we want negative three times 12, 5th is negative 36 5th plus six is negative 6/5 negative three times one. Fifth is negative 3/5 plus negative nine is negative 48. 5th and negative three times two is negative six plus 15 gives us a positive nine. Okay, now we have got the first column. All set and we can work on the second column. We want a one In this second column. 2nd row, so we're replacing row two. And how do we get that? 2/5 to become a one. We can multiply that row by its reciprocal. So multiplying that row by five halves. So five halves times row two. Because we're replacing road to we can copy over rose one and three. So one 12 5th. So 1/5 20 negative 6/5 negative 48 5th and nine. Now doing the work to replace rho to zero times five halves is 0, 2/5 times five halves is the one that we want. Ye 16 5th times five halves while the fives cancel out. So 16 divided by two is eight and negative 10 times five halves. Well that'll reduce to negative five times five which is negative 25. Okay. And now we want a zero below that one in the second column. So we want this negative 6/5 to become a zero, replacing row three. And in order to get that to be a zero. We work with the road that has the one in the desired position which is road to. We're going to multiply row two times 6/ and then We will add that to row three. Because we're replacing row three, we can copy over rows one and two. So that'll be 1, 12 5th 1/ 2018 negative 25. And doing the work to replace row 3 6/5 times 00 plus zero is 0, 6/5 times one is 6/5 plus a negative 6/5 is the zero that we want 6/5 times eight is going to be a 48 fifth scoops, that doesn't go there. 48 5th and then plus the negative 48 5th. Well that is zero interesting and 6/5 times negative 25 is going to be a negative 30 plus the nine is going to be a negative 21 interesting. We look at this final row and we remember that that corresponds to an equation for our system of equations. This is zero, X. Is plus zero wise plus zero Z. S equals negative 21. And there are just no values that we can plug in for X. Y and Z. To make that true. So this is impossible. When we come up with something that is impossible. Then we have no solution. We look at our answer choices and this matches with answer choice. D Well done. We'll catch you on the next one.

In Exercises 1 - 24, use Gaussian Eliminaion to find the complete solution to each system of equations, or show that none exists.5x + 12y + z = 102x + 5y + 2z = - 1x + 2y - 3z = 5

Key Concepts

Gaussian Elimination

Row Echelon Form

Consistency of a System

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